Question

A survey of 1,562 randomly selected adults showed that 522 of them have heard of a new electronic reader. The accompanying technology display results from a test of the claim that 35​% of adults have heard of the new electronic reader. Use the normal distribution as an approximation to the binomial​ distribution, and assume a 0.05 significance level to complete parts​ (a) through​ (e). Is the test two-tailed, left-tailed, or right-tailed? Left-tailed test Two-tailed test Right tailed test What is the test statistic? (Round to two décimal places as needed.) What is the P-vahie?

Answer 1

Answer:

a) We want to test the claim that 35​% of adults have heard of the new electronic reader, then the system of hypothesis are.:  

Null hypothesis:$p=0.35$  

Alternative hypothesis:$p \neq 0.35$  

And is a two tailed test

b) $z=\frac{0.334 -0.35}{\sqrt{\frac{0.35(1-0.35)}{1562}}}=-1.326$  

c) $p_v =2*P(z$  

d) Null hypothesis:$p=0.35$  

e) Fail to reject the null hypothesis because the P-value is greater than the significance level, alpha.

Step-by-step explanation:

Information provided

n=1562 represent the random sample selected

X=522 represent the people who have heard of a new electronic reader

$\hat p=\frac{522}{1562}=0.334$ estimated proportion of people who have heard of a new electronic reader

$p_o=0.35$ is the value to verify

$\alpha=0.05$ represent the significance level

z would represent the statistic

$p_v$ represent the p value

Part a

We want to test the claim that 35​% of adults have heard of the new electronic reader, then the system of hypothesis are.:  

Null hypothesis:$p=0.35$  

Alternative hypothesis:$p \neq 0.35$  

And is a two tailed test

Part b

The statistic for this case is given :

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

Replacing the info given we got:

$z=\frac{0.334 -0.35}{\sqrt{\frac{0.35(1-0.35)}{1562}}}=-1.326$  

Part c

We can calculate the p value using the laternative hypothesis with the following probability:

$p_v =2*P(z$  

Part d

The null hypothesis for this case would be:

Null hypothesis:$p=0.35$  

Part e

The best conclusion for this case would be:

Fail to reject the null hypothesis because the P-value is greater than the significance level, alpha.